1. **Problem Statement:** We have a continuous random variable $x$ with probability density function (pdf): $$f(x) = \begin{cases} \frac{1}{10} + e^{-2x+3}, & -3 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases}$$ We need to find the probabilities: (i) $P(x < 1)$, (ii) $P(x > 4)$, (iii) $P(x > 1)$, (iv) $P(x < -4)$, (v) $P(-2 \leq x \leq 2.5)$. 2. **Key Formula:** For a continuous random variable with pdf $f(x)$, the probability that $x$ lies between $a$ and $b$ is: $$P(a \leq x \leq b) = \int_a^b f(x) \, dx$$ 3. **Important Notes:** - Since $f(x) = 0$ outside $[-3,3]$, probabilities outside this range are 0 or 1 depending on the interval. - We will integrate $f(x)$ over the required intervals. 4. **Calculate each probability:** (i) $P(x < 1) = P(-3 \leq x < 1)$ since $f(x)=0$ for $x < -3$. $$P(x < 1) = \int_{-3}^1 \left( \frac{1}{10} + e^{-2x+3} \right) dx$$ Split the integral: $$= \int_{-3}^1 \frac{1}{10} dx + \int_{-3}^1 e^{-2x+3} dx$$ Calculate each: $$\int_{-3}^1 \frac{1}{10} dx = \frac{1}{10} (1 - (-3)) = \frac{4}{10} = 0.4$$ For the exponential: $$\int e^{-2x+3} dx = -\frac{1}{2} e^{-2x+3} + C$$ Evaluate from $-3$ to $1$: $$= -\frac{1}{2} \left( e^{-2(1)+3} - e^{-2(-3)+3} \right) = -\frac{1}{2} (e^{1} - e^{9}) = \frac{1}{2} (e^{9} - e^{1})$$ Numerical values: $e^{1} \approx 2.7183$, $e^{9} \approx 8103.08$ So: $$\frac{1}{2} (8103.08 - 2.7183) = \frac{1}{2} (8100.3617) = 4050.18085$$ This is very large, which suggests the pdf is not normalized. But we proceed as per problem. So, $$P(x < 1) = 0.4 + 4050.18085 = 4050.58085$$ (ii) $P(x > 4)$: Since $4 > 3$ and $f(x) = 0$ for $x > 3$, $$P(x > 4) = 0$$ (iii) $P(x > 1) = P(1 < x \leq 3)$ $$= \int_1^3 \left( \frac{1}{10} + e^{-2x+3} \right) dx = \int_1^3 \frac{1}{10} dx + \int_1^3 e^{-2x+3} dx$$ Calculate: $$\int_1^3 \frac{1}{10} dx = \frac{1}{10} (3-1) = \frac{2}{10} = 0.2$$ For exponential: $$= -\frac{1}{2} (e^{-2(3)+3} - e^{-2(1)+3}) = -\frac{1}{2} (e^{-3} - e^{1}) = \frac{1}{2} (e^{1} - e^{-3})$$ Numerical: $e^{1} \approx 2.7183$, $e^{-3} \approx 0.0498$ So: $$\frac{1}{2} (2.7183 - 0.0498) = \frac{1}{2} (2.6685) = 1.33425$$ Therefore, $$P(x > 1) = 0.2 + 1.33425 = 1.53425$$ (iv) $P(x < -4)$: Since $-4 < -3$ and $f(x) = 0$ for $x < -3$, $$P(x < -4) = 0$$ (v) $P(-2 \leq x \leq 2.5)$: $$= \int_{-2}^{2.5} \left( \frac{1}{10} + e^{-2x+3} \right) dx = \int_{-2}^{2.5} \frac{1}{10} dx + \int_{-2}^{2.5} e^{-2x+3} dx$$ Calculate: $$\int_{-2}^{2.5} \frac{1}{10} dx = \frac{1}{10} (2.5 - (-2)) = \frac{1}{10} (4.5) = 0.45$$ Exponential: $$= -\frac{1}{2} (e^{-2(2.5)+3} - e^{-2(-2)+3}) = -\frac{1}{2} (e^{-2} - e^{7}) = \frac{1}{2} (e^{7} - e^{-2})$$ Numerical: $e^{7} \approx 1096.633$, $e^{-2} \approx 0.1353$ So: $$\frac{1}{2} (1096.633 - 0.1353) = \frac{1}{2} (1096.4977) = 548.24885$$ Therefore, $$P(-2 \leq x \leq 2.5) = 0.45 + 548.24885 = 548.69885$$ **Final answers:** (i) $P(x < 1) \approx 4050.58$ (ii) $P(x > 4) = 0$ (iii) $P(x > 1) \approx 1.53$ (iv) $P(x < -4) = 0$ (v) $P(-2 \leq x \leq 2.5) \approx 548.70$ Note: The probabilities exceed 1, indicating the given function is not a valid pdf unless normalized. But these are the integrals as requested.